Chemistry - Thermochemistry Question with Solution | TestHub

The reaction is: $2{\mathrm{NO}}_2(\mathrm{g})\to {\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})$

Given:

$\Delta U_f^{\circ }[{\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})]=2kcal/mol$

$\Delta U_{reaction}^{\circ }=-16kcal/mol$

Temperature $T=727^{\circ }\mathrm{C}=1000\mathrm{K}$

We need to calculate $\Delta H_{formation}^{\circ }$ of ${\mathrm{NO}}_2$ (in kcal/mol).

First, calculate $\Delta U_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]$ using the given reaction:

$\Delta U_{reaction}^{\circ }=\sum \Delta U_f^{\circ }(\mathrm{products})-\sum \Delta U_f^{\circ }(\mathrm{reactants})$

$-16kcal/mol=\Delta U_f^{\circ }[{\mathrm{N}}_2{\mathrm{O}}_4(\mathrm{g})]-2\Delta U_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]$

$-16=2-2\Delta U_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]$

$2\Delta U_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]=2+16=18$

$\Delta U_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]=9kcal/mol$

Now, consider the formation reaction of ${\mathrm{NO}}_2$:

$\frac{1}{2}{\mathrm{N}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\to {\mathrm{NO}}_2(\mathrm{g})$

The relationship between enthalpy and internal energy is $\Delta H=\Delta U+\Delta n_{g}RT$.

For the formation of ${\mathrm{NO}}_2$:

$\Delta n_g=(molesofgaseousproducts)-(molesofgaseousreactants)$

$\Delta n_g=1-(\frac{1}{2}+1)=1-1.5=-0.5$

Using the ideal gas constant $R=2cal/(\mathrm{mol}\cdot \mathrm{K})=0.002kcal/(\mathrm{mol}\cdot \mathrm{K})$:

$\Delta H_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]=\Delta U_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]+\Delta n_{g}RT$

$\Delta H_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]=9kcal/mol+(-0.5)\times (0.002kcal/(\mathrm{mol}\cdot \mathrm{K}))\times (1000\mathrm{K})$

$\Delta H_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]=9-0.5\times 2$

$\Delta H_f^{\circ }[{\mathrm{NO}}_2(\mathrm{g})]=9-1=8kcal/mol$

The final answer is $\boxed{8}$.